proving an inequality involving a linear spline / piecewise polynomial I have $n+1$ sample points $x_i = \left(\frac{i}{n}\right)^4$ and want to approximate the function $f(x)=\sqrt{x}$ by a linear spline $f_n \in S^{1,0}(\mathcal{T_n})$ on the interval $[0,1]$. I know from a theorem that this spline is unique and can be written as
$$
f_n (x) = \sum_i y_i \phi_i(x)
$$
where $y_i = f(x_i)$ and $\phi_i(x)$ are certain hat functions which objey $\phi_i(x_j) = \delta_{ij}$ and form a basis of the spline space.
What I want to prove is the following:
$$
\max\limits_{x \in \left[0,1\right]} |f_n - f| \leq \frac{C}{n^2}
$$
where $C > 0$ and independent of $n$.
I know an estimate which involves the second derivative of $f$ but for $\sqrt{x}$ this diverges so I don't think that I can apply it. Any suggestions how I can proceed? I tried to use that the $\phi_i$ are bounded by $1$ but this didn't lead me to a sensible esimate.
 A: I recommend a direct comparison between the target $T$ and the approximation $A$. For instance, on the first interval $[0,h_1]$ the error $T-A$ is given by
$$ g(t) = \sqrt{th_1} - t \sqrt{h_1} = (\sqrt{t} - t)\sqrt{h_1}, \quad t \in [0,1].$$
A short calculation establishes that $|g(t)|\leq C \sqrt{h_1}$, but fortunately $h_1 = n^{-4}$, so we can survive the extra square root. I would try a direct comparison on the other subintervals as well.
EDIT: The origin of $g$. Any point $x$ on the interval $[0,h_1]$ can be written as $x=th_1$ for $t \in [0,1]$. Our target is $T = \sqrt{x} = \sqrt{th_1}$. Our approximation $A$ is the value of the spline at the point $x$, i.e. a weighted linear combination of $\sqrt{0}$ and $\sqrt{h_1}$. Specifically, $A = (1-t) \sqrt{0} + t \sqrt{h_1}$. Therefore 
$$ T - A = \sqrt{th_1} - t \sqrt{h_1}.$$
In reality, I did a linear transformation of the interval $[0,h_1]$ into $[0,1]$. 
In general, a interpolation polynomial of degree $1$ has this form $$ p(x) = \left(\frac{x-b}{a-b}\right) f(a) + \left(\frac{x-a}{b-a}\right) f(b). $$
Note that $$\frac{x-b}{a-b} + \frac{x-a}{b-a}=1,$$
which allows us to set $$t =\frac{x-a}{b-a}$$ and write
$$ \tilde{p}(t) = p(x(t)) = (1-t) f(a) + tf(b),$$
effectively transforming the problem to the interval $[0,1]$. Naturally, we must transform $f$ as well and compare $\tilde{p}$ with
$$\tilde{f}(t) = f(x(t)).$$
Personally, I like the representation afford by $\tilde{p}$ because it stresses the fact that I am connecting two points in the plane with a straight line. 
